, 2021

Solution:
The main claim is that if the light pulse reflects vertically (on the left/right edges) $a$ times and horizontally $b$ times, then $\gcd(a+1, b+1)=1$, and the number of regions is $\frac{(a+2)(b+2)}{2}$. This claim can be conjectured by looking at small values of $a$ and $b$; we give a full proof at the end.

Assuming the claim, we are trying to find the least possible value of $a+b$ when $(a+2)(b+2)=2 \cdot 2021 = 2 \cdot 43 \cdot 47$. This happens when $(a+2, b+2) = (47, 86)$, which also satisfies $\gcd(a+1, b+1)=1$, and gives $a+b = 47 + 86 - 4 = 129$.

We now prove the claim. Imagine that at each reflection, it is the square that gets reflected instead. Then the path $p$ of the light pulse becomes a straight segment $s$ from $(0,0)$ to $(a+1, b+1)$ of slope $+m=\frac{a+1}{b+1}$.

- The square starts as 1 region; the light pulse hitting a corner at the end creates 1 more region.
- Each reflection of the light pulse creates a region. These correspond to intersections of $s$ with a line $x=n$ or $y=n$ for $x \in [a], y \in **$. There are $a+b$ such intersections.
- Each self-intersection of $p$ creates a region. An intersection on $p$ corresponds to two on $s$, and each intersection of $s$ happens with a line of slope $-m$ passing through an even integral point, i.e. a line of the form $(b+1)x + (a+1)y = 2k$. The open segment $s$ intersects these lines for $k \in [ab + a + b]$. However, the $a+b$ intersections that happen on a gridline $x \in \mathbb{Z}$ or $y \in \mathbb{Z}$ do not count, so here we have an additional $ab/2$ regions.

Therefore, the total number of regions is
$$2 + a + b + \frac{ab}{2} = \frac{(a+2)(b+2)}{2}$$